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• Quaternion
  1,007 bytes (155 words) - 20:47, 14 October 2013
• 2020 AMC 10A Problems
  <cmath>\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}</cmath> What is the sum of all real numbers <math>x</math> for which <math>|x^2-12x+34|=2?</math>
  13 KB (1,968 words) - 18:32, 29 February 2024
• Asymptote: Logical Operators and Loops
  if (floor((i-j)/2)==((i-j)/2)) if (floor((i-j)/2)==((i-j)/2))
  2 KB (324 words) - 16:50, 2 October 2016
• Pigeonhole Principle/Solutions
  ...riends, then the remaining friends must have from <math>1</math> to <math>n-2</math> friends for the remaining friends not to also have no friends. By p ...same remainder when divided by 5. Since there are 5 possible remainders (0-4), by the pigeonhole principle, at least two of the integers must share the
  10 KB (1,617 words) - 01:34, 26 October 2021
• 2002 USAMO Problems/Problem 3
  ...it is necessary and sufficient that <math> P(x) = Q(x) + \prod_{i=1}^{n}(x-x_i) </math>. ...s. But a polynomial with real coefficients must have an even number of non-real roots, so <math>P(x) </math> must have <math>n </math> real roots. Sim
  4 KB (688 words) - 13:38, 4 July 2013
• 2007 AIME I Problems/Problem 10
  [[Image:AIME I 2007-10.png]] ...ath> (<math>j < k</math>) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that <math>j\ne k</math> to solve the p
  13 KB (2,328 words) - 00:12, 29 November 2023
• 2004 USAMO Problems
  ...egers <math> x, y \in S </math>, if <math> x+y \in S </math>, then <math> x-y \in S </math>. * [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamo-test.shtml 2004 USAMO Problems]
  3 KB (474 words) - 09:20, 14 May 2021
• 2007 AIME II Problems/Problem 3
  ...ince <math>\triangle ABE \cong \triangle CDF</math> and are both <math>5-12-13</math> [[right triangle]]s, we can come to the conclusion that the two ne A slightly more analytic/brute-force approach:
  6 KB (933 words) - 00:05, 8 July 2023
• 2007 AIME II Problems/Problem 13
  {{AIME box|year=2007|n=II|num-b=12|num-a=14}}
  3 KB (600 words) - 11:10, 22 January 2023
• 2007 AIME II Problems
  <div style="text-align:center;">[[Image:2007 AIME II-3.png]]</div> ..._{i+1}</math> if <math>a_{i}</math> is [[even]]. How many four-digit parity-monotonic integers are there?
  9 KB (1,435 words) - 01:45, 6 December 2021
• 2003 USAMO Problems/Problem 1
  ...-1} + a </math> is divisible by 5, which is true when <math> k \equiv -3^{n-1}a \pmod{5} </math>. Since there is an odd digit in each of the residue cl ...s after it, and all multiples of 5 end in 5. Therefore, <math> a*10^x*5^{n-x}</math> always contains a 5 as its <math> (x+1)^{st}</math> digit, and we
  4 KB (736 words) - 22:17, 3 March 2023
• 2003 USAMO Problems/Problem 3
  ...{} c_j = \cdots = c_{j+i} = \cdots = c_k = j </math> (<math> 0 \le i \le k-j</math>). ''Proof.'' Since the <math>j </math> terms <math>c_0, \ldots, c_{j-1} </math> are all less than <math>j </math>, no other terms that precede <m
  3 KB (636 words) - 13:39, 4 July 2013
• Pentagon
  1 KB (191 words) - 09:59, 6 June 2022
• 2004 IMO Shortlist Problems/N2
  ...e-i}) = p^e + \sum_{i=0}^{e-1}p^i(p^{e-i} - p^{e-i-1}) = p^e + e(p^e - p^{e-1}) </math>. </center> ...^{e_i} - e_i \cdot p_i^{e_i}]}{\prod p_i^{e_i}} = \prod \left(e_i \frac{p_i-1}{p_i} + 1 \right) </math>.
  6 KB (1,007 words) - 09:10, 29 August 2011
• Parallelepiped
  2 KB (248 words) - 20:08, 17 August 2023
• Chicken McNugget Theorem
  ...am + bn</math> for [[nonnegative]] integers <math>a, b</math> is <math>mn-m-n</math>. ...e proof is based on the fact that in each pair of the form <math>(k, mn-m-n-k)</math>, exactly one element is expressible.
  17 KB (2,748 words) - 19:22, 24 February 2024
• Dodecagon
  A '''dodecagon''' is a 12-sided [[polygon]]. The sum of its internal [[angle]]s is <math>1800^{\circ}<
  1 KB (219 words) - 13:08, 15 June 2018
• Triangular number
  <math>1 +2+3 + 4............. +(n-1)+(n)</math>. ...math>T_{n} = 1 + 2 + \ldots + (n-1) + n = (1 + 2 + \ldots + n-1) + n = T_{n-1} + n</math>.
  2 KB (275 words) - 08:39, 7 July 2021
• 2007 AMC 12A Problems/Problem 24
  ...mod{4}</math> is that <math>v_2(2j+1)=0,</math> so we must have <math>v_2(n-1)>v_2(n+1)</math> since <math>v_2(2k) \geq 1 >v_2(2j+1).</math> Therefore, {{AMC12 box|year=2007|num-b=23|num-a=25|ab=A}}
  4 KB (588 words) - 14:40, 23 August 2023
• User talk:1=2
  * [[1962 AHSME]] (Complete '''w/o solutions w/o problems 41-50''')
  13 KB (1,464 words) - 17:28, 6 January 2024
• 2006 AIME II Problems/Problem 13
  ...math>, <math>n\geq 0</math>. Then the final odd integer is <math>2n+1 + 2(j-1) = 2(n+j) - 1</math>. The odd integers form an [[arithmetic sequence]] wit ...ual <math>0</math>. We then perform casework based on the parity of <math>p-q</math>.
  4 KB (675 words) - 10:40, 14 July 2022
• 2006 AIME II Problems
  When rolling a certain unfair six-sided die with faces numbered <math>1, 2, 3, 4, 5</math>, and <math>6</math> ...h> and <math>t_2</math> intersect at <math>(x,y),</math> and that <math>x=p-q\sqrt{r},</math> where <math>p, q,</math> and <math>r</math> are positive i
  8 KB (1,350 words) - 12:00, 4 December 2022
• User:Temperal/The Problem Solver's Resource4
  ==<span style="font-size:20px; color: blue;">Algebra</span>== <cmath>f(x)=a_nx^n+a_{n-1}x^{n-1}\ldots+a_0</cmath>, where <math>a_n\ne 0</math>, and <math>a_i</math> are
  4 KB (828 words) - 21:45, 27 February 2020
• 1998 PMWC Problems
  for(int j = n-i; j > 0; --j){ ...ad a total of <math>30</math> balls, how many balls were there in the right-hand box?
  11 KB (1,738 words) - 19:25, 10 March 2015
• 2001 PMWC Problems
  F=O+(G-O)+(E-O); draw((xstart,A.y)--(xstart,A.y-len));
  4 KB (641 words) - 21:24, 21 April 2014
• 2002 PMWC Problems
  label("$" + string(i) + "^\textrm{" + ord[i-1] + "}$ Row:", (-5,(-i+1)*sqrt(3)/2+correction)); draw( (1-cos(pi/3)*correction,-sin(pi/3)*correction)--(0+cos(pi/3)*correction,-sqrt(3
  5 KB (725 words) - 16:07, 23 April 2014
• 2003 PMWC Problems
  filldraw((p-a-b)--(p+a-b)--(p+a+b)--(p-a+b)--cycle,black); fill((p-a-b)--(p+a-b)--(p+a+b)--(p-a+b)--cycle,white);
  7 KB (918 words) - 16:15, 22 April 2014
• 1989 USAMO Problems/Problem 3
  Let <math>P(z)= z^n + c_1 z^{n-1} + c_2 z^{n-2} + \cdots + c_n</math> be a [[polynomial]] in the complex variable <math>z <cmath> \lvert z_i-i \rvert \cdot \lvert z_j-i \rvert < 1. </cmath>
  2 KB (340 words) - 19:11, 18 July 2016
• Cramer's Rule
  2 KB (352 words) - 18:22, 11 October 2023
• Lucas' Theorem
  ...n of <math>n</math> in [[base]] <math>p</math> and <math>(\overline{i_mi_{m-1}\cdots i_0})_p</math> is the representation of <math>i</math> in base <mat For all <math>1\leq k \leq p-1</math>, <math>\binom{p}{k}\equiv 0 \pmod{p}</math>. Then we have
  1 KB (251 words) - 15:13, 11 August 2020
• Summation
  ...r algebraic structure), <math>\sum_{i=a}^{b}c_i=c_a+c_{a+1}+c_{a+2}...+c_{b-1}+c_{b}</math>. Here <math>i</math> refers to the index of summation, <math ...i= \frac{n(n+1)}{2}</math>, and in general <math>\sum_{i=a}^{b} i= \frac{(b-a+1)(a+b)}{2}</math>
  3 KB (482 words) - 16:39, 8 October 2023
• Zorn's Lemma
  ...intermediate results, viz., Bourbaki's Theorem (also known as the Bourbaki-Witt theorem). ...h>x \in A</math> does <math>x = f(x)</math>, which contradicts the Bourbaki-Witt Theorem. {{halmos}}
  9 KB (1,669 words) - 19:02, 1 August 2018
• Ideal
  ...ls in rings are the [[kernel]]s of ring [[homomorphism]]s; in this way, two-sided ideals of rings are similar to [[normal subgroup]]s of [[group]]s. ...frak{a}</math> is both a left ideal and a right ideal, it is called a ''two-sided ideal''. In a [[commutative ring]], all three kinds of ideals are the
  8 KB (1,389 words) - 23:44, 17 February 2020
• Work
  1 KB (192 words) - 00:46, 16 July 2018
• Celsius
  4 KB (357 words) - 22:40, 17 April 2024
• 1999 USAMO Problems/Problem 4
  <cmath> \sum_{i=k+1}^n -a_i \le 2k-n . </cmath> <cmath> \sum_{i=k+1}^n a_i^2 \le \left( \sum_{i=k+1}^n-a_i \right)^2 \le (2k-n)^2 . </cmath>
  3 KB (499 words) - 09:47, 20 July 2016
• 2001 IMO Shortlist Problems/A3
  By the [[Cauchy-Schwarz Inequality]], By the [[Cauchy-Schwarz Inequality]]
  3 KB (441 words) - 00:54, 24 October 2023
• 1999 IMO Problems/Problem 2
  ...ath> and the right-hand side becomes <math>(x_a + x_b)^4/8</math>. By [[AM-GM]], ...ase the left-hand side of the desired inequality without changing the right-hand side; and then to use the inductive hypothesis.
  4 KB (838 words) - 01:04, 17 November 2023
• Proofs of AM-GM
  This pages lists some proofs of the weighted [[AM-GM Inequality]]. The inequality's statement is as follows: for all nonnegati ...and <math>\lambda_i \neq 0</math>, for some <math>i</math>, then the right-hand side of the inequality is zero and the left hand of the inequality is g
  12 KB (2,171 words) - 07:55, 11 May 2023
• Carleman's Inequality
  Define <math>c_n = n \left( 1 + \frac{1}{n} \right)^n = \frac{(n+1)^n}{n^{n-1}}</math>. Then for all positive integers <math>k</math>, Now, by [[AM-GM]],
  2 KB (278 words) - 16:39, 29 December 2021
• 2008 AMC 12B Problems/Problem 23
  ...square. When these are multiplied, they equal <math>2^{a+n-a} \times 5^{b+n-b} = 10^n</math>. <math>\log 10^n=n</math> so the number of factors divided .../math> satisfies <math>ab = 10^n</math> -- ex. <math>(1, 10^n), (2, 5*10^{n-1})</math>, etc. Then the sum of the base-<math>10</math> logarithms is <mat
  5 KB (814 words) - 18:02, 17 January 2023
• Vieta's formulas
  Let <math>P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0</math> be any polynomial with [[Complex number | c ....</cmath> This can be compactly summarized as <math>s_j = (-1)^j \frac{a_{n-j}}{a_n}</math> for some <math>j</math> such that <math>1 \leq j \leq n</mat
  3 KB (512 words) - 15:31, 22 February 2024
• 2000 AMC 12 Problems/Problem 16
  <math>f(i, j) = 17(i-1) + j = 17i + j - 17</math> {{AMC12 box|year=2000|num-b=15|num-a=17}}
  2 KB (310 words) - 11:28, 3 August 2021
• 2007 AMC 10A Problems
  draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle); ...at random from 0 to 2007, inclusive. What is the probability that <math>ad-bc</math> is even?
  13 KB (2,058 words) - 17:54, 29 March 2024
• 1990 USAMO Problems/Problem 4
  '''Lemma.''' Let <math>a</math> be an integer. Then there are <math>2^{k-1}</math> <math>k</math>-good sequences starting on <math>a</math>, and furt ...M</math>. Then the only possibilities for <math>a_{k+1}</math> are <math>m-1</math> and <math>M+1</math>; either way, <math>\{ a_i \}_{i=1}^{k+1}</math
  3 KB (529 words) - 19:15, 18 July 2016
• 2007 AMC 10A Problems/Problem 18
  ...\cdot 12 = \frac {36}5 </math>. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.co {{AMC10 box|year=2007|ab=A|num-b=17|num-a=19}}
  6 KB (867 words) - 00:17, 20 May 2023
• 1974 USAMO Problems/Problem 1
  <cmath> a-b \mid P(a) - P(b) = b-c \mid P(b)-P(c) = c-a \mid P(c) - P(a) = a-b , </cmath> ...absolute value. In fact, two of them, say <math>(a-b)</math> and <math>(b-c)</math>, must be equal. Then
  7 KB (1,291 words) - 20:30, 27 April 2020
• 1994 USAMO Problems
  The sides of a 99-gon are initially colored so that consecutive sides are red, blue, red, blue ...sual, let <math>\, \binom{n}{k} \,</math> denote <math>\, n! \over k! \, (n-k)!</math>. Prove that <cmath> \sum_{U \subseteq S} (-1)^{|U|} \binom{m - \s
  2 KB (391 words) - 07:58, 19 July 2016
• Isogonal conjugate
  Let <math>I'</math> and <math>J'</math> be the A-excenters of triangles <math>\triangle ABC</math> and <math>\triangle ADC,</ ...oordinates <cmath>(p' : q' : r'), p' = \frac {|B - C|^2}{p}, q' = \frac {|A-C|^2}{q}, r' = \frac {|A - B|^2}{r}.</cmath>
  54 KB (9,416 words) - 08:40, 18 April 2024
• 1976 IMO Problems/Problem 2
  ...inct real solutions, where <math>2^{n-1}</math> are positive and <math>2^{n-1}</math> are negative. Also, for ever root <math>r</math>, <math>|r|<2</mat ...values less than 2, where <math>2^{k-1}</math> are positive and <math>2^{k-1}</math> are negative.
  3 KB (596 words) - 16:19, 28 July 2015
• 1976 IMO Problems/Problem 5
  {{IMO box|year=1976|num-b=4|num-a=6}}
  2 KB (377 words) - 16:28, 29 January 2021
• Partition (combinatorics)
  ...iven by <math>F(x)= \sum_{n \geq 0}P(n) x^n = \prod_{n=1}^\infty \frac{1}{1-x^n}</math>. Partitions can also be studied by using the [[Jacobi theta func ...epresents a different addend in the partition. The rows are ordered in non-increasing order so that that the row with the most dots is on the top and t
  10 KB (1,508 words) - 14:24, 17 September 2017
• 2001 AMC 12 Problems/Problem 6
  A telephone number has the form <math>\text{ABC-DEF-GHIJ}</math>, where each letter represents {{AMC12 box|year=2001|num-b=5|num-a=7}}
  2 KB (340 words) - 03:02, 28 June 2023
• 2008 AIME II Problems
  ...ch is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin bikin a^2 + y^2 = b^2 + x^2 = (a-x)^2 + (b-y)^2
  7 KB (1,167 words) - 21:33, 12 August 2020
• 2008 AIME I Problems/Problem 3
  <math>(13,1)</math> and <math>(3,9)</math> give non-integral <math>b</math>, but <math>(8,5)</math> gives <math>b = 15</math>. T ...<math>2y+3z+4x=91</math>. Subtracting the two expressions we get <math>y+z-2x=-17</math>. Note that <math>-17</math> is odd, so one of <math>x,y,z</mat
  3 KB (564 words) - 22:15, 28 November 2023
• Mock AIME 5 Pre 2005 Problems
  Two 5-digit numbers are called "responsible" if they are:
  6 KB (909 words) - 07:27, 12 October 2022
• 2008 AIME II Problems/Problem 10
  ...our points using coordinates <math>0 \le x,y \le 3</math>, with the bottom-left point being <math>(0,0)</math>. By the [[Pythagorean Theorem]], the dis {{AIME box|year=2008|n=II|num-b=9|num-a=11}}
  4 KB (569 words) - 09:44, 25 November 2019
• 2005 USAMO Problems/Problem 2
  (1) <math>(x^3+y)(x^3+1) = (x^3+y)(x+1)(x^2-x+1) = 147^{157} = 7^{314}3^{152}</math>, <math>x^2-x+1 = (x+1)^2-3(x+1)+3 \rightarrow \gcd(x+1, x^2-x+1) \le 3</math>.
  7 KB (1,053 words) - 10:38, 12 August 2015
• 2005 USAMO Problems/Problem 1
  ...t with the circular arrangement of <math>n,p_{1}p_{2},p_{2}p_{3}\ldots,p_{k-1}p_{k}</math> as shown. ...by placing <math>p_k,p^{2}_{k},\ldots,p^{e_k}_{k}</math> between <math>p_{k-1}p_k</math> and <math>n</math>. It is easy to see that each element of <mat
  4 KB (650 words) - 13:40, 4 July 2013
• 2006 IMO Problems/Problem 5
  a_1 - a_0 &\mid P(a_1)- P(a_0) = a_2-a_1 \\ &\mid P(a_r)- P(a_{r-1}) = a_1 - a_0 .
  3 KB (704 words) - 14:42, 7 September 2021
• Schreier's Theorem
  ...iscovered it in 1928, and used it to give an improved proof of the [[Jordan-Hölder Theorem]]. Six years later, Hans Zassenhaus published his [[Zassenh ...ies in question. For integers <math>j \in [1,m-1]</math>, <math>i \in [0,n-1]</math>, let <math>H'_{im+j} = H_{i+1} \cdot (H_i \cap K_j)</math>, and fo
  2 KB (337 words) - 12:13, 9 April 2019
• 2008 Mock ARML 2 Problems
  ...th>f(x)</math> equal the number of zeroes to the right of the rightmost non-zero digit in the decimal form of <math>x!</math>, and let <math>n = \frac { ...sum_{i = 0}^{n}a_ia_{n - i} = 1</math> and <math>a_n > 0</math> for all non-negative integers <math>n</math>, evaluate <math>\sum_{j = 0}^{\infty}\frac
  4 KB (582 words) - 21:57, 8 May 2019
• 2008 Mock ARML 2 Problems/Problem 8
  ...sum_{i = 0}^{n}a_ia_{n - i} = 1</math> and <math>a_n > 0</math> for all non-negative integers <math>n</math>, evaluate <math>\sum_{j = 0}^{\infty}\frac By the given, the coefficients on the right-hand side are all equal to <math>1</math>, yielding the [[geometric series]]
  1 KB (190 words) - 00:57, 31 May 2016
• 2003 AMC 12A Problems/Problem 17
  <math>x^2 + (y-4)^2 = 4^2</math> <math>(x-2)^2 + y^2 = 2^2</math>
  6 KB (1,026 words) - 22:35, 29 March 2023
• 2003 AMC 12B Problems/Problem 18
  ...s 7 and 11 in order to keep <math>x</math> at a minimum. Moving all the non-y terms to the left hand side of the equation, we end up with: <cmath>7^{5b+1}11^{5d-1}=y^{13}</cmath>
  6 KB (914 words) - 11:07, 7 September 2023
• Jean-Victor Poncelet
  '''Jean-Victor Poncelet''' (July 1, 1788 – December 22, 1867) was a French mathema ...son of Claude de Poncelet, a lawyer in the local Parliament, and a well-to-do landowner. Poncelet attended local schools in his home town before going
  2 KB (253 words) - 11:41, 19 December 2018
• KGS math club/solution 6 1
  ...know that <math>\forall i: 2 | (W-w_i)</math>, that is, each remaining ball-mass is divisible by two. Combining these, we get <math>\forall i,j: 2 | (w_i-w_j)</math> by subtracting the case i from the case j.
  4 KB (759 words) - 06:27, 18 July 2009
• Legendre's Formula
  ...i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}</cmath> <cmath>e_2(27!)=\frac{27-S_2(27)}{2-1}=27-S_2(27)</cmath>
  4 KB (699 words) - 17:55, 5 August 2023
• 2001 IMO Shortlist Problems/C2
  ...that <math>S(x)\equiv y\bmod{n}</math> for all integers <math>0\leq y\leq n-1</math>. This means that the sum of every <math>S(a)</math> is congruent to ...(c_i\sum_{j=1}^{n} (n-1)!*a_j\right)=\sum_{i=1}^{n} c_i\cdot \frac{n(n+1)(n-1)!}{2}=\frac{c_i(n+1)!}{2}\bmod{n!}</cmath>
  2 KB (364 words) - 08:55, 31 August 2011
• 2001 IMO Shortlist Problems/C5
  519 bytes (73 words) - 20:07, 29 May 2020
• 2001 IMO Shortlist Problems/N6
  ...ath>(c+d-b)^2+b^2\equiv c^2+d^2\pmod p</math>. Factorizing, <cmath>2(b^2-bc-bd+cd)\equiv0\pmod p</cmath> <cmath>\implies(b-c)(b-d)\equiv0\pmod p</cmath>
  2 KB (443 words) - 13:08, 17 August 2011
• Millennium Problems
  ...], the [[Poincaré Conjecture]], the [[Riemann Hypothesis]], and the [[Yang-Mills Theory]]. In 2003, the Poincaré Conjecture was proven by Russian math ...ent. He gave three lectures, titled "Ricci Flow and Geometrization of Three-Manifolds", on April 7, 9, and 11. These were his first public presentation
  13 KB (1,969 words) - 17:57, 22 February 2024
• 1989 AHSME Problems/Problem 29
  {{AHSME box|year=1989|num-b=28|num-a=30}}
  1 KB (173 words) - 19:52, 30 December 2020
• 2008 IMO Problems/Problem 5
  a_i-n, & \mbox{ if } a_i \in B .../math> elements from <math>\cal{N}</math>, which would imply <math>N = 2^{k-n}M</math> and solve the problem.
  4 KB (677 words) - 01:10, 19 November 2023
• 2000 USAMO Problems
  Call a real-valued function <math>f</math> very convex if
  3 KB (495 words) - 19:02, 18 April 2014
• 2000 USAMO Problems/Problem 4
  if((i == 0 || j == 9) && !(j-i == 9)) fill(shift(i,j)*unitsquare,rgb(0.3,0.3,0.3)); ...r of the rows containing the colored squares. Hence, each of the <math>1999-m</math> colored squares must be placed in different rows, but as there are
  2 KB (382 words) - 13:37, 4 July 2013
• 2000 USAMO Problems/Problem 5
  ...e externally tangent at the point <math>A_k</math>, so <math>O_k, A_k, O_{k-1}</math> are [[collinear]]. ...ath>\theta_k = 180 - \theta_{k-1}</math>, and so <math>\theta_k = \theta_{k-2}</math>. Thus, <math>\theta_{1} = \theta_{7}</math>.
  3 KB (609 words) - 09:52, 20 July 2016
• 2001 USAMO Problems/Problem 1
  ...olors in 3 boxes and <math>3n-48</math> colors in 2. Thus <math>n\geq 20+48-2n,</math> so <math>3n\geq 68</math>, and <math>n\geq23</math> and we are do {{USAMO newbox|year=2001|before=First question|num-a=2}}
  5 KB (841 words) - 17:19, 5 May 2022
• Karamata's Inequality
  ...a\leq x_1\leq x_2 \leq b</math> and <math>\Gamma(x, y):=\frac{f(y)-f(x)}{y-x}</math>, <math>\Gamma(x_1, x)\leq \Gamma (x_2, x)</math> <cmath>\Gamma(x_1, x)=\frac{f(x_1)-f(x)}{x_1-x}\leq \frac{f(x_2)-f(x)}{x_2-x}=\Gamma(x_2, x)</cmath>
  2 KB (370 words) - 03:39, 28 March 2024
• 1973 USAMO Problems/Problem 2
  <center> <math>X_0=1</math>, <math>X_1=1</math>, <math>X_{n+1}=X_n+2X_{n-1}</math> <math>(n=1,2,3,\dots),</math></center> <center><math>Y_0=1</math>, <math>Y_1=7</math>, <math>Y_{n+1}=2Y_n+3Y_{n-1}</math> <math>(n=1,2,3,\dots)</math>.</center>
  2 KB (342 words) - 18:54, 3 July 2013
• 2004 IMO Problems/Problem 4
  ...math>t_3 = t_2 + t_1 + k</math> for <math>k \ge 0</math>; then (by [[Cauchy-Schwarz Inequality]]) ...} \right) + t_n \sum_{i=1}^{n-1} \frac 1{t_i} + \frac{1}{t_n} \sum_{i=1}^{n-1} t_i + 1\end{align*}</cmath>
  2 KB (322 words) - 00:54, 19 November 2023
• 2002 AMC 10A Problems
  Given that <math>a, b,</math> and <math>c</math> are non-zero real numbers, define <math>(a, b, c) = \frac{a}{b} + \frac{b}{c} + \fra ..., and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
  11 KB (1,733 words) - 11:04, 12 October 2021
• 2002 AMC 10A Problems/Problem 20
  {{AMC10 box|year=2002|ab=A|num-b=19|num-a=21}}
  3 KB (439 words) - 22:15, 9 June 2023
• 1973 USAMO Problems/Problem 1
  http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1973Problem1 {{USAMO box|year=1973|before=First Question|num-a=2}}
  2 KB (276 words) - 08:10, 4 January 2022
• 1975 USAMO Problems/Problem 3
  ...], we can write <math>Q(x)</math> as <cmath>Q(x) = c(x)(x-1)(x-2) \cdots (x-n)</cmath> where <math>c</math> is a constant. Thus, <cmath>(x+1)P(x) - x = c(x)(x-1)(x-2) \cdots (x-n).</cmath>
  3 KB (465 words) - 17:29, 8 January 2024
• 1962 IMO Problems/Problem 7
  <math>M_{12}=\frac{V_1-V_2}{2}=\left( \frac{\sqrt{2}}{6}r_c ,\;\frac{\sqrt{6}}{6}r_c,\;-\frac{r_c}{ <math>M_{13}=\frac{V_1-V_3}{2}=\left( \frac{\sqrt{2}}{6}r_c ,\;-\frac{\sqrt{6}}{6}r_c,\;-\frac{r_c}
  11 KB (1,928 words) - 20:52, 21 November 2023
• 1963 IMO Problems/Problem 3
  ...textrm{-coordinate} = \sum_{k = 1}^{\frac{n}{2}}a_k \cdot \sin \frac{2\pi(k-1)}{n}.\textbf{ (1)}</cmath> ...xtrm{-coordinate} = \sum_{k = \frac{n}{2}+1}^{n}a_k \cdot \sin \frac{2\pi(k-1)}{n}</cmath>
  4 KB (836 words) - 17:58, 7 December 2022
• 2001 AMC 10 Problems
  two-digit number such that <math>N = P(N)+S(N)</math>. What is the units digit o ...t), and the rest sells for half price. How much money is raised by the full-price tickets?
  14 KB (1,983 words) - 16:25, 2 June 2022
• 2000 AMC 10 Problems/Problem 16
  ..., hence we can compute the slope, <math>m_1</math> to be <math>\frac{0-3}{6-0}=-\frac{1}{2}</math>, so <math>l_1</math> is the line <math>y=\frac{-1}{2} ...th> gives us <math>b_2=-2</math>, so <math>l_2</math> is the line <math>y=x-2</math>.
  7 KB (966 words) - 23:22, 31 July 2023
• 2007 IMO Shortlist Problems/A1
  <center><math>\max\{|x_i-a_i|:1\leq i\leq n\}\geq \frac{d}{2}</math>.</center> ..._j:i\le j\le n\}</math>, all <math>d_i</math> can be expressed as <math>a_p-a_q</math>, where <math>1\le p\le i\le q \le n</math>.
  3 KB (734 words) - 05:11, 26 March 2024
• 2004 AMC 12B Problems/Problem 14
  pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) );
  7 KB (1,083 words) - 22:41, 23 November 2020
• 1976 IMO Problems
  For all positive integral <math>n</math>, <math>u_{n+1}=u_n(u_{n-1}^2-2)-u_1</math>, <math>u_0=2</math>, and <math>u_1=2\frac12</math>. Prove that
  2 KB (409 words) - 16:25, 29 January 2021
• 1971 IMO Problems/Problem 6
  ...})(i, j = 1, 2, \cdots, n)</math> be a square matrix whose elements are non-negative integers. Suppose that whenever an element <math>a_{ij} = 0</math>, ...h>. We thus infer that<cmath>s(\ell_j)+s(c_{\sigma(j)})\ge (n-t)+(n-k+t)=2n-k\ge n</cmath>so <math>s(\ell_j)+s(c_{\sigma(j)})\ge n,\ \forall j\ge k+1</m
  6 KB (1,192 words) - 14:14, 29 January 2021
• 2001 AMC 12 Problems/Problem 10
  {{AMC12 box|year=2001|num-b=9|num-a=11}} {{AMC10 box|year=2001|num-b=17|num-a=19}}
  2 KB (252 words) - 00:11, 15 August 2022
• 2001 AMC 12 Problems/Problem 22
  Therefore <math>[EHJ]=[ACD]-[AHD]-[DEH]-[EJC]=35-14-\frac {21}2-\frac{15}2 = \boxed{3}</math>. {{AMC12 box|year=2001|num-b=21|num-a=23}}
  7 KB (1,120 words) - 02:27, 29 August 2023
• 2009 AMC 12B Problems
  Each morning of her five-day workweek, Jane bought either a <math>50</math>-cent muffin or a <math>75 Twenty percent less than <math>60</math> is one-third more than what number?
  13 KB (2,030 words) - 03:04, 5 September 2021
• 2009 AMC 12B Problems/Problem 20
  ...rtices, implying each vertex owns 4 edge ends. There are twice as many edge-ends as there are edges, and <math>2 \cdot 100 = 200</math>. {{AMC12 box|year=2009|ab=B|num-b=19|num-a=21}}
  9 KB (1,567 words) - 13:43, 19 August 2023
• 2009 AMC 12B Problems/Problem 25
  Given any square, we can circumscribe another axes-parallel square around it. In the picture below, the original square is red ...n a clever bijection given in [http://web2.slc.qc.ca/sh/Contest/AMC12_2009B-S.pdf this page].
  15 KB (2,229 words) - 03:36, 4 September 2021
• Maximal ideal
  968 bytes (183 words) - 19:50, 23 August 2009
• Polynomial ring
  ...biggr)\cdot \biggl(\sum_j b_jx^j\biggr) = \sum_k\biggl(\sum_{i=0}^k a_ib_{k-i}\biggr)x^k ...nto <math>R[x]/(x-a)</math> and the canonical homomorphism of <math>R[x]/(x-a)</math> into <math>R</math>.)
  12 KB (2,010 words) - 00:10, 3 August 2020

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